The concept of a numerical sequence implies that each natural number corresponds to some real value. Such a series of numbers can be both arbitrary and have certain properties - a progression. In the latter case, each subsequent element (member) of the sequence can be calculated using the previous one.

An arithmetic progression is a sequence of numerical values ​​in which its adjacent members differ by the same number(all elements of the series, starting from the 2nd one, have a similar property). This number - the difference between the previous and subsequent member - is constant and is called the progression difference.

Progression Difference: Definition

Consider a sequence consisting of j values ​​A = a(1), a(2), a(3), a(4) … a(j), j belongs to the set natural numbers N. An arithmetic progression, according to its definition, is a sequence in which a(3) - a(2) = a(4) - a(3) = a(5) - a(4) = ... = a(j) – a(j-1) = d. The value of d is the desired difference of this progression.

d = a(j) - a(j-1).

Allocate:

• An increasing progression, in which case d > 0. Example: 4, 8, 12, 16, 20, …
• decreasing progression, then d< 0. Пример: 18, 13, 8, 3, -2, …

Difference of progression and its arbitrary elements

If 2 arbitrary members of the progression (i-th, k-th) are known, then the difference for this sequence can be established based on the relation:

a(i) = a(k) + (i - k)*d, so d = (a(i) - a(k))/(i-k).

The progression difference and its first term

This expression will help determine the unknown value only in cases where the number of the sequence element is known.

Progression difference and its sum

The sum of a progression is the sum of its members. To calculate the total value of its first j elements, use the corresponding formula:

S(j) =((a(1) + a(j))/2)*j, but since a(j) = a(1) + d(j – 1), then S(j) = ((a(1) + a(1) + d(j – 1))/2)*j=(( 2a(1) + d(– 1))/2)*j.

Arithmetic and geometric progressions

Theoretical information

Theoretical information

	Arithmetic progression	Geometric progression
Definition	Arithmetic progression a n a sequence is called, each member of which, starting from the second, is equal to the previous member, added with the same number d (d- progression difference)	geometric progression b n a sequence of non-zero numbers is called, each term of which, starting from the second, is equal to the previous term multiplied by the same number q (q- denominator of progression)
Recurrent formula	For any natural n a n + 1 = a n + d	For any natural n b n + 1 = b n ∙ q, b n ≠ 0
nth term formula	a n = a 1 + d (n - 1)	b n \u003d b 1 ∙ q n - 1, b n ≠ 0
characteristic property
Sum of the first n terms

Examples of tasks with comments

Exercise 1

AT arithmetic progression (a n) a 1 = -6, a 2

According to the formula of the nth term:

a 22 = a 1+ d (22 - 1) = a 1+ 21d

By condition:

a 1= -6, so a 22= -6 + 21d.

It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = - 48.

Answer : a 22 = -48.

Task 2

Find the fifth term of the geometric progression: -3; 6;....

1st way (using n-term formula)

According to the formula of the n-th member of a geometric progression:

b 5 \u003d b 1 ∙ q 5 - 1 = b 1 ∙ q 4.

Because b 1 = -3,

2nd way (using recursive formula)

Since the denominator of the progression is -2 (q = -2), then:

b 3 = 6 ∙ (-2) = -12;

b 4 = -12 ∙ (-2) = 24;

b 5 = 24 ∙ (-2) = -48.

Answer : b 5 = -48.

Task 3

In arithmetic progression ( a n) a 74 = 34; a 76= 156. Find the seventy-fifth term of this progression.

For an arithmetic progression, the characteristic property has the form .

Therefore:

.

Substitute the data in the formula:

Answer: 95.

Task 4

In arithmetic progression ( a n ) a n= 3n - 4. Find the sum of the first seventeen terms.

To find the sum of the first n terms of an arithmetic progression, two formulas are used:

.

Which one in this case more convenient to use?

By condition, the formula of the nth member of the original progression is known ( a n) a n= 3n - 4. Can be found immediately and a 1, and a 16 without finding d . Therefore, we use the first formula.

Answer: 368.

Task 5

In arithmetic progression a n) a 1 = -6; a 2= -8. Find the twenty-second term of the progression.

According to the formula of the nth term:

a 22 = a 1 + d (22 – 1) = a 1+ 21d.

By condition, if a 1= -6, then a 22= -6 + 21d. It is necessary to find the difference of progressions:

d= a 2 – a 1 = -8 – (-6) = -2

a 22 = -6 + 21 ∙ (-2) = -48.

Answer : a 22 = -48.

Task 6

Several consecutive terms of a geometric progression are recorded:

Find the term of the progression, denoted by the letter x .

When solving, we use the formula for the nth term b n \u003d b 1 ∙ q n - 1 for geometric progressions. The first member of the progression. To find the denominator of the progression q, you need to take any of these terms of the progression and divide by the previous one. In our example, you can take and divide by. We get that q \u003d 3. Instead of n, we substitute 3 in the formula, since it is necessary to find the third term of a given geometric progression.

Substituting the found values ​​into the formula, we get:

.

Answer : .

Task 7

From the arithmetic progressions given by the formula of the nth term, choose the one for which the condition is satisfied a 27 > 9:

Since the specified condition must be satisfied for the 27th term of the progression, we substitute 27 instead of n in each of the four progressions. In the 4th progression we get:

.

Answer: 4.

Task 8

In arithmetic progression a 1= 3, d = -1.5. Specify highest value n , for which the inequality a n > -6.

Arithmetic progression name a sequence of numbers (members of a progression)

In which each subsequent term differs from the previous one by a steel term, which is also called step or progression difference.

Thus, by setting the step of the progression and its first term, you can find any of its elements using the formula

Properties of an arithmetic progression

1) Each member of the arithmetic progression, starting from the second number, is the arithmetic mean of the previous and next member of the progression

The converse is also true. If the arithmetic mean of neighboring odd (even) members of the progression is equal to the member that stands between them, then this sequence of numbers is an arithmetic progression. By this assertion it is very easy to check any sequence.

Also by the property of arithmetic progression, the above formula can be generalized to the following

This is easy to verify if we write the terms to the right of the equal sign

It is often used in practice to simplify calculations in problems.

2) The sum of the first n terms of an arithmetic progression is calculated by the formula

Remember well the formula for the sum of an arithmetic progression, it is indispensable in calculations and is quite common in simple life situations.

3) If you need to find not the entire sum, but a part of the sequence starting from its k -th member, then the following sum formula will come in handy in you

4) It is of practical interest to find the sum of n members of an arithmetic progression starting from the kth number. To do this, use the formula

This is where the theoretical material ends and we move on to solving problems that are common in practice.

Example 1. Find the fortieth term of the arithmetic progression 4;7;...

Solution:

According to the condition, we have

Define the progression step

According to the well-known formula, we find the fortieth term of the progression

Example2. The arithmetic progression is given by its third and seventh members. Find the first term of the progression and the sum of ten.

Solution:

We write the given elements of the progression according to the formulas

We subtract the first equation from the second equation, as a result we find the progression step

The found value is substituted into any of the equations to find the first term of the arithmetic progression

Calculate the sum of the first ten terms of the progression

Without applying complex calculations, we found all the required values.

Example 3. An arithmetic progression is given by the denominator and one of its members. Find the first term of the progression, the sum of its 50 terms starting from 50, and the sum of the first 100.

Solution:

Let's write the formula for the hundredth element of the progression

and find the first

Based on the first, we find the 50th term of the progression

Finding the sum of the part of the progression

and the sum of the first 100

The sum of the progression is 250.

Example 4

Find the number of members of an arithmetic progression if:

a3-a1=8, a2+a4=14, Sn=111.

Solution:

We write the equations in terms of the first term and the step of the progression and define them

We substitute the obtained values ​​​​into the sum formula to determine the number of members in the sum

Making simplifications

and solve the quadratic equation

Of the two values ​​found, only the number 8 is suitable for the condition of the problem. Thus the sum of the first eight terms of the progression is 111.

Example 5

solve the equation

1+3+5+...+x=307.

Solution: This equation is the sum of an arithmetic progression. We write out its first term and find the difference of the progression

What is the essence of the formula?

This formula allows you to find any BY HIS NUMBER" n" .

Of course, you need to know the first term a 1 and progression difference d, well, without these parameters, you can’t write down a specific progression.

It is not enough to memorize (or cheat) this formula. It is necessary to assimilate its essence and apply the formula in various problems. Yes, and do not forget at the right time, yes ...) How not forget- I do not know. But how to remember If needed, I'll give you a hint. For those who master the lesson to the end.)

So, let's deal with the formula of the n-th member of an arithmetic progression.

What is a formula in general - we imagine.) What is an arithmetic progression, a member number, a progression difference - is clearly stated in the previous lesson. Take a look if you haven't read it. Everything is simple there. It remains to figure out what nth member.

progression in general view can be written as a series of numbers:

a 1 , a 2 , a 3 , a 4 , a 5 , .....

a 1- denotes the first term of an arithmetic progression, a 3- third member a 4- fourth, and so on. If we are interested in the fifth term, let's say we are working with a 5, if one hundred and twentieth - from a 120.

How to define in general any member of an arithmetic progression, s any number? Very simple! Like this:

a n

That's what it is n-th member of an arithmetic progression. Under the letter n all the numbers of members are hidden at once: 1, 2, 3, 4, and so on.

And what does such a record give us? Just think, instead of a number, they wrote down a letter ...

This notation gives us a powerful tool for working with arithmetic progressions. Using the notation a n, we can quickly find any member any arithmetic progression. And a bunch of tasks to solve in progression. You will see further.

In the formula of the nth member of an arithmetic progression:

a n = a 1 + (n-1)d

a 1- the first member of the arithmetic progression;

n- member number.

The formula links the key parameters of any progression: a n ; a 1 ; d and n. Around these parameters, all the puzzles revolve in progression.

The nth term formula can also be used to write a specific progression. For example, in the problem it can be said that the progression is given by the condition:

a n = 5 + (n-1) 2.

Such a problem can even confuse ... There is no series, no difference ... But, comparing the condition with the formula, it is easy to figure out that in this progression a 1 \u003d 5, and d \u003d 2.

And it can be even angrier!) If we take the same condition: a n = 5 + (n-1) 2, yes, open the brackets and give similar ones? We get a new formula:

an = 3 + 2n.

it Only not general, but for a specific progression. This is where the pitfall lies. Some people think that the first term is a three. Although in reality the first member is a five ... A little lower we will work with such a modified formula.

In tasks for progression, there is another notation - a n+1. This is, you guessed it, the "n plus the first" term of the progression. Its meaning is simple and harmless.) This is a member of the progression, the number of which is greater than the number n by one. For example, if in some problem we take for a n fifth term, then a n+1 will be the sixth member. Etc.

Most often the designation a n+1 occurs in recursive formulas. Do not be afraid of this terrible word!) This is just a way of expressing a term of an arithmetic progression through the previous one. Suppose we are given an arithmetic progression in this form, using the recurrent formula:

a n+1 = a n +3

a 2 = a 1 + 3 = 5+3 = 8

a 3 = a 2 + 3 = 8+3 = 11

The fourth - through the third, the fifth - through the fourth, and so on. And how to count immediately, say the twentieth term, a 20? But no way!) While the 19th term is not known, the 20th cannot be counted. This is the fundamental difference between the recursive formula and the formula of the nth term. Recursive works only through previous term, and the formula of the nth term - through the first and allows straightaway find any member by its number. Not counting the whole series of numbers in order.

In an arithmetic progression, a recursive formula can easily be turned into a regular one. Count a pair of consecutive terms, calculate the difference d, find, if necessary, the first term a 1, write the formula in the usual form, and work with it. In the GIA, such tasks are often found.

Application of the formula of the n-th member of an arithmetic progression.

To begin, consider direct application formulas. At the end of the previous lesson there was a problem:

Given an arithmetic progression (a n). Find a 121 if a 1 =3 and d=1/6.

This problem can be solved without any formulas, simply based on the meaning of the arithmetic progression. Add, yes add ... An hour or two.)

And according to the formula, the solution will take less than a minute. You can time it.) We decide.

The conditions provide all the data for using the formula: a 1 \u003d 3, d \u003d 1/6. It remains to be seen what n. No problem! We need to find a 121. Here we write:

Please pay attention! Instead of an index n a specific number appeared: 121. Which is quite logical.) We are interested in the member of the arithmetic progression number one hundred twenty one. This will be our n. It is this meaning n= 121 we will substitute further into the formula, in brackets. Substitute all the numbers in the formula and calculate:

a 121 = 3 + (121-1) 1/6 = 3+20 = 23

That's all there is to it. Just as quickly one could find the five hundred and tenth member, and the thousand and third, any. We put instead n the desired number in the index of the letter " a" and in brackets, and we consider.

Let me remind you the essence: this formula allows you to find any term of an arithmetic progression BY HIS NUMBER" n" .

Let's solve the problem smarter. Let's say we have the following problem:

Find the first term of the arithmetic progression (a n) if a 17 =-2; d=-0.5.

If you have any difficulties, I will suggest the first step. Write down the formula for the nth term of an arithmetic progression! Yes Yes. Hand write, right in your notebook:

a n = a 1 + (n-1)d

And now, looking at the letters of the formula, we understand what data we have and what is missing? Available d=-0.5, there is a seventeenth member ... Everything? If you think that's all, then you can't solve the problem, yes ...

We also have a number n! In the condition a 17 =-2 hidden two options. This is both the value of the seventeenth member (-2) and its number (17). Those. n=17. This "little thing" often slips past the head, and without it, (without the "little thing", not the head!) The problem cannot be solved. Although ... and without a head too.)

Now we can just stupidly substitute our data into the formula:

a 17 \u003d a 1 + (17-1) (-0.5)

Oh yes, a 17 we know it's -2. Okay, let's put it in:

-2 \u003d a 1 + (17-1) (-0.5)

That, in essence, is all. It remains to express the first term of the arithmetic progression from the formula, and calculate. You get the answer: a 1 = 6.

Such a technique - writing a formula and simply substituting known data - helps a lot in simple tasks. Well, you must, of course, be able to express a variable from a formula, but what to do!? Without this skill, mathematics can not be studied at all ...

Another popular problem:

Find the difference of the arithmetic progression (a n) if a 1 =2; a 15 =12.

What are we doing? You will be surprised, we write the formula!)

a n = a 1 + (n-1)d

Consider what we know: a 1 =2; a 15 =12; and (special highlight!) n=15. Feel free to substitute in the formula:

12=2 + (15-1)d

Let's do the arithmetic.)

12=2 + 14d

d=10/14 = 5/7

This is the correct answer.

So, tasks a n , a 1 and d decided. It remains to learn how to find the number:

The number 99 is a member of an arithmetic progression (a n), where a 1 =12; d=3. Find the number of this member.

We substitute the known quantities into the formula of the nth term:

a n = 12 + (n-1) 3

At first glance, there are two unknown quantities here: a n and n. But a n is some member of the progression with the number n... And this member of the progression we know! It's 99. We don't know his number. n, so this number also needs to be found. Substitute the progression term 99 into the formula:

99 = 12 + (n-1) 3

We express from the formula n, we think. We get the answer: n=30.

And now a problem on the same topic, but more creative):

Determine if the number 117 will be a member of an arithmetic progression (a n):

-3,6; -2,4; -1,2 ...

Let's write the formula again. What, there are no options? Hm... Why do we need eyes?) Do we see the first member of the progression? We see. This is -3.6. You can safely write: a 1 \u003d -3.6. Difference d can be determined from the series? It's easy if you know what the difference of an arithmetic progression is:

d = -2.4 - (-3.6) = 1.2

Yes, we did the simplest thing. It remains to deal with an unknown number n and an incomprehensible number 117. In the previous problem, at least it was known that it was the term of the progression that was given. But here we don’t even know that ... How to be!? Well, how to be, how to be... Turn on Creative skills!)

We suppose that 117 is, after all, a member of our progression. With an unknown number n. And, just like in the previous problem, let's try to find this number. Those. we write the formula (yes-yes!)) and substitute our numbers:

117 = -3.6 + (n-1) 1.2

Again we express from the formulan, we count and get:

Oops! The number turned out fractional! One hundred and one and a half. And fractional numbers in progressions can not be. What conclusion do we draw? Yes! Number 117 is not member of our progression. It is somewhere between the 101st and 102nd members. If the number turned out to be natural, i.e. positive integer, then the number would be a member of the progression with the found number. And in our case, the answer to the problem will be: no.

Task based on a real version of the GIA:

The arithmetic progression is given by the condition:

a n \u003d -4 + 6.8n

Find the first and tenth terms of the progression.

Here the progression is set in an unusual way. Some kind of formula ... It happens.) However, this formula (as I wrote above) - also the formula of the n-th member of an arithmetic progression! She also allows find any member of the progression by its number.

We are looking for the first member. The one who thinks. that the first term is minus four, is fatally mistaken!) Because the formula in the problem is modified. The first term of an arithmetic progression in it hidden. Nothing, we'll find it now.)

Just as in the previous tasks, we substitute n=1 into this formula:

a 1 \u003d -4 + 6.8 1 \u003d 2.8

Here! The first term is 2.8, not -4!

Similarly, we are looking for the tenth term:

a 10 \u003d -4 + 6.8 10 \u003d 64

That's all there is to it.

And now, for those who have read up to these lines, the promised bonus.)

Suppose, in a difficult combat situation of the GIA or the Unified State Exam, you forgot the useful formula of the n-th member of an arithmetic progression. Something comes to mind, but somehow uncertainly ... Whether n there, or n+1, or n-1... How to be!?

Calm! This formula is easy to derive. Not very strict, but to be sure and right decision that's enough!) For the conclusion, it is enough to remember the elementary meaning of the arithmetic progression and have a couple of minutes of time. You just need to draw a picture. For clarity.

We draw a numerical axis and mark the first one on it. second, third, etc. members. And note the difference d between members. Like this:

We look at the picture and think: what is the second term equal to? Second one d:

a 2 =a 1 + 1 d

What is the third term? Third term equals first term plus two d.

a 3 =a 1 + 2 d

Do you get it? I don't put some words in bold for nothing. Okay, one more step.)

What is the fourth term? Fourth term equals first term plus three d.

a 4 =a 1 + 3 d

It's time to realize that the number of gaps, i.e. d, always one less than the number of the member you are looking for n. That is, up to the number n, number of gaps will be n-1. So, the formula will be (no options!):

a n = a 1 + (n-1)d

In general, visual pictures are very helpful in solving many problems in mathematics. Don't neglect the pictures. But if it's difficult to draw a picture, then ... only a formula!) In addition, the formula of the nth term allows you to connect the entire powerful arsenal of mathematics to the solution - equations, inequalities, systems, etc. You can't put a picture in an equation...

Tasks for independent decision.

For warm-up:

1. In arithmetic progression (a n) a 2 =3; a 5 \u003d 5.1. Find a 3 .

Hint: according to the picture, the problem is solved in 20 seconds ... According to the formula, it turns out more difficult. But for mastering the formula, it is more useful.) In Section 555, this problem is solved both by the picture and by the formula. Feel the difference!)

And this is no longer a warm-up.)

2. In arithmetic progression (a n) a 85 \u003d 19.1; a 236 =49, 3. Find a 3 .

What, reluctance to draw a picture?) Still! It's better in the formula, yes ...

3. Arithmetic progression is given by the condition:a 1 \u003d -5.5; a n+1 = a n +0.5. Find the one hundred and twenty-fifth term of this progression.

In this task, the progression is given in a recurrent way. But counting up to the one hundred and twenty-fifth term... Not everyone can do such a feat.) But the formula of the nth term is within the power of everyone!

4. Given an arithmetic progression (a n):

-148; -143,8; -139,6; -135,4, .....

Find the number of the smallest positive term of the progression.

5. According to the condition of task 4, find the sum of the smallest positive and largest negative members of the progression.

6. The product of the fifth and twelfth terms of an increasing arithmetic progression is -2.5, and the sum of the third and eleventh terms is zero. Find a 14 .

Not the easiest task, yes ...) Here the method "on the fingers" will not work. You have to write formulas and solve equations.

Answers (in disarray):

3,7; 3,5; 2,2; 37; 2,7; 56,5

Happened? It's nice!)

Not everything works out? It happens. By the way, in the last task there is one subtle point. Attentiveness when reading the problem will be required. And logic.

The solution to all these problems is discussed in detail in Section 555. And the fantasy element for the fourth, and the subtle moment for the sixth, and general approaches for solving any problems for the formula of the nth term - everything is painted. I recommend.

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

Before we start to decide arithmetic progression problems, consider what a number sequence is, since an arithmetic progression is a special case of a number sequence.

A number sequence is a number set, each element of which has its own serial number . The elements of this set are called members of the sequence. The ordinal number of a sequence element is indicated by an index:

The first element of the sequence;

The fifth element of the sequence;

- "nth" element of the sequence, i.e. the element "standing in the queue" at number n.

There is a dependency between the value of a sequence element and its ordinal number. Therefore, we can consider a sequence as a function whose argument is the ordinal number of an element of the sequence. In other words, one can say that the sequence is a function of the natural argument:

The sequence can be specified in three ways:

1 . The sequence can be specified using a table. In this case, we simply set the value of each member of the sequence.

For example, Someone decided to do personal time management, and to begin with, to calculate how much time he spends on VKontakte during the week. By writing the time in a table, he will get a sequence consisting of seven elements:

The first line of the table contains the number of the day of the week, the second - the time in minutes. We see that, that is, on Monday Someone spent 125 minutes on VKontakte, that is, on Thursday - 248 minutes, and, that is, on Friday, only 15.

2 . The sequence can be specified using the nth member formula.

In this case, the dependence of the value of a sequence element on its number is expressed directly as a formula.

For example, if , then

To find the value of a sequence element with a given number, we substitute the element number into the formula for the nth member.

We do the same if we need to find the value of a function if the value of the argument is known. We substitute the value of the argument instead in the equation of the function:

If, for example, , then

Once again, I note that in a sequence, in contrast to an arbitrary numeric function, only a natural number can be an argument.

3 . The sequence can be specified using a formula that expresses the dependence of the value of the member of the sequence with number n on the value of the previous members. In this case, it is not enough for us to know only the number of a sequence member in order to find its value. We need to specify the first member or first few members of the sequence.

For example, consider the sequence ,

We can find the values ​​of the members of a sequence in sequence, starting from the third:

That is, each time to find the value of the nth member of the sequence, we return to the previous two. This way of sequencing is called recurrent, from the Latin word recurro- come back.

Now we can define an arithmetic progression. An arithmetic progression is a simple special case of a numerical sequence.

Arithmetic progression is called a numerical sequence, each member of which, starting from the second, is equal to the previous one, added with the same number.

The number is called the difference of an arithmetic progression. The difference of an arithmetic progression can be positive, negative, or zero.

If title="(!LANG:d>0">, то каждый член арифметической прогрессии больше предыдущего, и прогрессия является !} increasing.

For example, 2; 5; eight; eleven;...

If , then each term of the arithmetic progression is less than the previous one, and the progression is waning.

For example, 2; -one; -four; -7;...

If , then all members of the progression are equal to the same number, and the progression is stationary.

For example, 2;2;2;2;...

The main property of an arithmetic progression:

Let's look at the picture.

We see that

, and at the same time

Adding these two equalities, we get:

.

Divide both sides of the equation by 2:

So, each member of the arithmetic progression, starting from the second, is equal to the arithmetic mean of two neighboring ones:

Moreover, since

, and at the same time

, then

, and hence

Each member of the arithmetic progression starting with title="(!LANG:k>l">, равен среднему арифметическому двух равноотстоящих. !}

th member formula.

We see that for the members of the arithmetic progression, the following relations hold:

and finally

We got formula of the nth term.

IMPORTANT! Any member of an arithmetic progression can be expressed in terms of and . Knowing the first term and the difference of an arithmetic progression, you can find any of its members.

The sum of n members of an arithmetic progression.

In an arbitrary arithmetic progression, the sums of terms equally spaced from the extreme ones are equal to each other:

Consider an arithmetic progression with n members. Let the sum of n members of this progression be equal to .

Arrange the terms of the progression first in ascending order of numbers, and then in descending order:

Let's pair it up:

The sum in each parenthesis is , the number of pairs is n.

We get:

So, the sum of n members of an arithmetic progression can be found using the formulas:

Consider solving arithmetic progression problems.

1 . The sequence is given by the formula of the nth member: . Prove that this sequence is an arithmetic progression.

Let us prove that the difference between two adjacent members of the sequence is equal to the same number.

We have obtained that the difference of two adjacent members of the sequence does not depend on their number and is a constant. Therefore, by definition, this sequence is an arithmetic progression.

2 . Given an arithmetic progression -31; -27;...

a) Find the 31 terms of the progression.

b) Determine if the number 41 is included in this progression.

a) We see that ;

Let's write down the formula for the nth term for our progression.

In general

In our case , that's why